Question

The correct set of values of $\text{n},\text{l},{\text{m}}_{\text{l}}$ and ${\text{m}}_{\text{s}}$ for the singly occupied $1s^1$ orbital is:

• A. $n$ = 1, $l$ = 0, $m_l$ = -1, $m_s$ = 0
• B. $n$ = 2, $l$ = 0, $m_l$ = -1, $m_s$ = 0
• C. $n$ = 1, $l$ = 1, $m_l$ = -1, $m_s$ = -1/2
• D. $n$ = 1, $l$ = 0, $m_l$ = 0, $m_s$ = + 1/2

Answer 1

The correct option is D $n$ = 1, $l$ = 0, $m_l$ = 0, $m_s$ = + 1/2

Each electron in an atom is described by a set of four quantum numbers $\text{n}$, $\text{l}$, ${\text{m}}_{\text{l}}$ and ${\text{m}}_{\text{s}}$.
The given configuration $1s^1$ states that the electron is in the first shell and 's' sub-shell, hence, the value of $\text{n}$ = 1 and $\text{l}$ = 0.
Since $\text{l}$ = 0, ${\text{m}}_{\text{l}}$ is also 0 and the spin of an electron could be either +1/2 or -1/2.

Thus, the correct set of quantum numbers is $n=1,l=0,m=0,s=+\frac{1}{2}$.